Question: Given that $\sin A+\sin B=1$ and $\cos A+\cos B= \frac{3}{2}$, what is the value of $\cos(A-B)$?
Answer: Squaring both equations, we get $\sin^2 A + 2 \sin A \sin B + \sin^2 B = 1$ and $\cos^2 A + 2 \cos A \cos B + \cos^2 B = \frac{9}{4},$ so
\[\sin^2 A + 2 \sin A \sin B + \sin^2 B + \cos^2 A + 2 \cos A \cos B + \cos^2 B = \frac{13}{4}.\]Then $2 \sin A \sin B + 2 \cos A \cos B = \frac{13}{4} - 2 = \frac{5}{4},$ so from the angle subtraction formula,
\[\cos (A - B) = \cos A \cos B + \sin A \sin B = \boxed{\frac{5}{8}}.\]